Question bank/JEE Main/Physics

A small smooth disc of mass m and radius r moving with an initial
velocity along the positive x-axis collided with a big disc of mass
2 m and radius 2 r which was initially at rest with its centre at origin
as show in figure.
[IMAGE] If the coefficient of restitution is 0 then velocity of larger disc
after collision is

A$\frac{8v}{27}\overset{\hat{}}{i} - \frac{2\sqrt{2}}{27}v\overset{\hat{}}{j}$B$\frac{8v}{27}\overset{\hat{}}{i} + \frac{2\sqrt{2}}{27}v\overset{\hat{}}{j}$C$\frac{v}{3}\overset{\hat{}}{i}$D$\frac{2\sqrt{2}}{27}v\overset{\hat{}}{i} - \frac{8v}{27}\overset{\hat{}}{j}$

Show answer and solution▾

[IMAGE] The larger sphere will move along line of impact.
e=0, magnitude of velocity of larger sphere
$v^{'} = \frac{mvcos\theta}{m + 2m} = \frac{vcos\theta}{3}$
Cos$\theta$= $2\sqrt{2}/3$
Sin $\theta$= 1/3
Velocity of larger sphere
$= v^{'}cos\theta\overset{\hat{}}{i} - v^{'}sin\theta\overset{\hat{}}{j}$
$= \frac{v}{3}\cos^{2}\theta\overset{\hat{}}{i} - \frac{v}{3}sin\theta cos\theta\overset{\hat{}}{j} = \frac{8}{27}v\overset{\hat{}}{i} - \frac{2\sqrt{2}}{27}v\overset{\hat{}}{j}$

Source: Grass-G / P-16 Physics paper 25 Nov First.docx